Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Complex Number

  1. #1
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1

    Complex Number

    Hi, my question is :
    Let n be a positive integer. Find all real numbers x such that (x-i)^n + (x+i)^n = 0
    I find that there is no such REAL number.
    Does someone find a different answer?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    Quote Originally Posted by vincisonfire View Post
    Hi, my question is :
    Let n be a positive integer. Find all real numbers x such that (x-i)^n + (x+i)^n = 0
    I find that there is no such REAL number.
    Does someone find a different answer?
    Hint: convert to polars and use De Moivre's Theorem.

    You should find that it reduces down to

    2(x^2 + 1)^{\frac{n}{2}}\cos{(nx)} = 0.

    You should be able to solve this with null factor law.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by vincisonfire View Post
    Hi, my question is :
    Let n be a positive integer. Find all real numbers x such that (x-i)^n + (x+i)^n = 0
    I find that there is no such REAL number.
    Does someone find a different answer?
    Well a pretty obvious solution from inspection is that if n is even, then x = 0 is a real number solution. But I'll do the working just to see if any others lurk:
     (x-i)^n = -(x+i)^n

     x-i = (-(x+i)^n)^{\frac{1}{n}}

     x-i = (-1)^{\frac{1}{n}} (x+i)

     x-i = \frac{(-1)^1}{(-1)^n} \times (x+i)

     x-i = \frac{-1}{(-1)^n} \times (x+i)

     x-i = \frac{-x}{(-1)^n} - \frac{i}{(-1)^n}

    Get all x terms on LHS and non-x terms on RHS

     x+ \frac{x}{(-1)^n} = i- \frac{i}{(-1)^n}

     x(1+ \frac{1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n}

     x(\frac{(-1)^n+1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n}

     x = \frac{i(-1)^n-i}{(-1)^n} \times \frac{(-1)^n}{(-1)^n+1}

     x = \frac{i(-1)^n-i}{(-1)^n+1}

     x = \frac{i((-1)^n-1)}{(-1)^n+1}

    Hence  n \neq 2k+1 for k any positive integer.

    For  n = 2k  x = 0

    Hence the equation has a real number solution for all even values of n. And that solution is  x = 0. 0 is a real number.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    Shouldn't it be <br /> <br />
2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0<br />
because the angle will be arctan\frac{1}{x}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    Quote Originally Posted by vincisonfire View Post
    Shouldn't it be <br /> <br />
2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0<br />
because the angle will be *arctan\frac{1}{x}
    Yes you're right, my apologies.

    I did \frac{\Re{z}}{\Im{z}} instead of \frac{\Im{z}}{\Re{z}}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    I would thus have  x = \frac{2n}{(2k+1)\pi}, k\in \mathbb Z but it does not work.
    And thanks Mush. I guess there is no solution for all n.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    the solutions are x=\cot \left(\frac{(2k+1)\pi}{2n} \right), \ 1 \leq k \leq n. just put x=\cot \alpha and solve a simple equation.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    Mush somewhere you did *(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n} *
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    Quote Originally Posted by vincisonfire View Post
    Mush somewhere you did *(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n} *
    I was wondering about that too.

    Shouldn't it be \sqrt[n]{-1}?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    To NonCommAlg :
    I don't understand why but as n goes up the solutions are not working???
    e.g. when n = 101 mathematica gives me 2.59828*10^167 + 0. I
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    Yes it should be <br /> <br />
\sqrt[n]{-1}<br />
. That complicates the things ... but the answer is still good.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by vincisonfire View Post
    To NonCommAlg :
    I don't understand why but as n goes up the solutions are not working???
    the solution still work but you won't get any new ones because: \cot \left(\frac{(2k+1)\pi}{2n} \right)=\cot \left(\frac{(2(k+n)+1)\pi}{2n} \right).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    No I meant (Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I
    When n > 10 it begins to have great values.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    OK this is driving me mad.

    According to my CAS, it reduces to...

    2\cos{\left(n\arctan{x} - \frac{n\pi}{2}\right)}(x^2 + 1)^{\frac{n}{2}}

    and gives

    x = -\cot{\left(\frac{k\pi}{n} - \frac{\pi}{2n}\right)} and -1 \leq \frac{n + 2k - 1}{n} \leq 1.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by vincisonfire View Post
    No I meant (Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I
    When n > 10 it begins to have great values.
    the solution i gave you is exact. you trust the software you're using more than mathematics?!!
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 2nd 2010, 02:54 PM
  2. Replies: 3
    Last Post: September 13th 2010, 12:13 PM
  3. Complex number
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 16th 2009, 07:59 AM
  4. complex number
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 29th 2008, 07:00 PM
  5. [SOLVED] Complex D Name F D Number
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 14th 2007, 09:51 AM

Search Tags


/mathhelpforum @mathhelpforum